Question

The length of a cylinder is measured with a meter rod having least count 0.1cm. Its diameter is measured with Vernier calipers having least count 0.01cm. Given that length is 5.0cm and radius is 2cm. Find the percentage error in the calculated value of the volume.

Answer 1

Hint
The vernier calipers found in the laboratory incorporates a main scale and a sliding vernier scale which allows readings to the nearest 0.02 mm. This instrument may be used to measure outer dimensions of objects (using the main jaws), inside dimensions (using the smaller jaws at the top), and depths (using the stem).

Complete step by step answer
The volume of a cylinder is $V = \pi {r^2}h$
According to the question length (h) and radius (r) given which are 5 cm and 2cm respectively
The percentage error in volume is $\dfrac{{\Delta V}}{V} \times 100$ which is equal to sum of twice the percentage error in radius and percentage error in length.
$\dfrac{{\Delta V}}{V} \times 100 = \dfrac{{2\Delta r}}{r} \times 100 + \dfrac{{\Delta h}}{h} \times 100$
Putting the values which are given in the question,
$ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = \dfrac{{2 \times 0.01}}{2} \times 100 + \dfrac{{0.1}}{5} \times 100$
$ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = (1 + 2) = 3\% $
So the percentage error in the volume is $= 3%$.

Note
To calculate percentage error, Subtract the accepted value from the experimental value. Divide that answer by the accepted value. Multiply that answer by 100 and add the % symbol to express the answer as a percentage.